Question: The equation of a circle $C$ is $x^2+y^2-2x-8y-8 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-2x) + (y^2-8y) = 8$ $(x^2-2x+1) + (y^2-8y+16) = 8 + 1 + 16$ $(x-1)^{2} + (y-4)^{2} = 25 = 5^2$ Thus, $(h, k) = (1, 4)$ and $r = 5$.